Answer:
[tex]2.8 m/s^2[/tex]
Explanation:
We are given that
Initial speed,u=8 m/s
Final speed,v=6 m/s
Distance,s=5 m
We have to find the magnitude of her acceleration as she shows.
We know that
[tex]v^2-u^2=2as[/tex]
Using the formula
[tex](6)^2-(8)^2=2a(5)=10a[/tex]
[tex]-28=10a[/tex]
[tex]a=\frac{-28}{10}=-2.8 m/s^2[/tex]
Magnitude of acceleration=[tex]\mid a\mid=2.8 m/s^2[/tex]
Answer:
Acceleration, a = [tex]2.8\ m/s^2[/tex]
Explanation:
Given that,
Initial speed of the skier, u = 8 m/s
Final speed of the skier, v = 6 m/s
Distance, d = 5 m
We need to find the magnitude of her acceleration. It is equal to the rate of change of speed. It is given by :
[tex]v^2-u^2=2as\\\\a=\dfrac{v^2-u^2}{2s}\\\\a=\dfrac{6^2-8^2}{2\times 5}\\\\a=-2.8\ m/s^2[/tex]
So, the magnitude of her acceleration is [tex]2.8\ m/s^2[/tex].
